If $f(x) =\cot ^{-1}\left(\frac{x^x-x^{-x}}{2}\right)$, then f'(1) equals

-1

We have,

$f(x)=\cot ^{-1}\left(\frac{x^x-x^{-x}}{2}\right)$

$\Rightarrow f^{\prime}(x)=-\frac{1}{1+\left(\frac{x^x-x^{-x}}{2}\right)^2} . \frac{d}{d x}\left(\frac{x^x-x^{-x}}{2}\right)$

$\Rightarrow f^{\prime}(x)=\frac{-2}{4+\left(x^x-x^{-x}\right)^2} . \frac{d}{d x}\left(x^x-x^{-x}\right)$

$\Rightarrow f^{\prime}(x)=\frac{-2}{\left(x^x+x^{-x}\right)^2} \frac{d}{d x}\left(e^{x \log x}-e^{-x \log x}\right)$

$\Rightarrow f^{\prime}(x)=\frac{-2}{\left(x^x+x^{-x}\right)^2} \times\left\{e^{x \log x} . \frac{d}{d x}(x \log x)-e^{-x \log x} \frac{d}{d x}(-x \log x)\right\}$

$\Rightarrow f^{\prime}(x)=\frac{-2}{\left(x^x+x^{-x}\right)^2}\left\{x^x(1+\log x)+x^{-x}(1+\log x)\right\}$

$\Rightarrow f^{\prime}(x)=\frac{-2(1+\log x)}{\left(x^x+x^{-x}\right)^2} .\left(x^x+x^{-x}\right)=\frac{-2(1+\log x)}{x+x^{-x}}$

$\Rightarrow f^{\prime}(1)=\frac{-2}{(1+1)}=-1$